Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $a = \dfrac{4p + 24}{p^2 + 2p - 24} \div \dfrac{p + 8}{5p - 20} $
Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{4p + 24}{p^2 + 2p - 24} \times \dfrac{5p - 20}{p + 8} $ First factor the quadratic. $a = \dfrac{4p + 24}{(p - 4)(p + 6)} \times \dfrac{5p - 20}{p + 8} $ Then factor out any other terms. $a = \dfrac{4(p + 6)}{(p - 4)(p + 6)} \times \dfrac{5(p - 4)}{p + 8} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ 4(p + 6) \times 5(p - 4) } { (p - 4)(p + 6) \times (p + 8) } $ $a = \dfrac{ 20(p + 6)(p - 4)}{ (p - 4)(p + 6)(p + 8)} $ Notice that $(p + 6)$ and $(p - 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 20(p + 6)\cancel{(p - 4)}}{ \cancel{(p - 4)}(p + 6)(p + 8)} $ We are dividing by $p - 4$ , so $p - 4 \neq 0$ Therefore, $p \neq 4$ $a = \dfrac{ 20\cancel{(p + 6)}\cancel{(p - 4)}}{ \cancel{(p - 4)}\cancel{(p + 6)}(p + 8)} $ We are dividing by $p + 6$ , so $p + 6 \neq 0$ Therefore, $p \neq -6$ $a = \dfrac{20}{p + 8} ; \space p \neq 4 ; \space p \neq -6 $